Research Article | Open Access

# Fixed Point Theorems of Quasicontractions on Cone Metric Spaces with Banach Algebras

**Academic Editor:**Simeon Reich

#### Abstract

We introduce the concept of quasicontractions on cone metric spaces with Banach algebras, and by a new method of proof, we will prove the existence and uniqueness of fixed points of such mappings. The main result generalizes the well-known theorem of Ćirić (Ćirić 1974).

#### 1. Introduction

Letbe a complete metric space. Recall that a mappingis called a quasicontraction if, for someand for all, one has Ćirić [1] introduced and studied quasicontractions as one of the most general classes of contractive-type mappings. He proved the well-known theorem that any quasicontractionhas a unique fixed point. Recently, scholars obtained various similar results on cone metric spaces. See, for instance, [2–5].

In this paper, we study the quasicontractions on metric spaces with Banach algebras, which are introduced in [6] and turn out to be an interesting generalization of classic metric spaces. By a new method of proof, we generalize Ćirić theorem.

Letalways be a real Banach algebra with a multiplication unit; that is,for all. An elementis said to be invertible if there is an inverse elementsuch that. The inverse ofis denoted by. For more details, we refer to [7].

The following proposition is well known (see [7]).

Proposition 1 (see [7]). *Letbe a Banach algebra with a unit, and let . If the spectral radiusofis less than 1, that is,
**
thenis invertible. Actually,
*

A subsetofis called a cone if (1)is nonempty closed and; (2)for all nonnegative real numbers; (3); (4).

For a given cone , we can define a partial orderingwith respect tobyif and only if . And will stand forand, whilewill stand for , where denotes the interior of.

*Remark 2. *In the literature on cone metric spaces, authors useto meanand andto mean. To our knowledge, and from a topological point of view, the order relationplays a very similar role in cone metric spaces asdoes in.

The coneis called normal if there is a numbersuch that for all, The least positive number satisfying above is called the normal constant of(see [8]).

In the following, we always assume thatis a cone inwith andis partial ordering with respect to.

*Definition 3 (see [8]). *Letbe a nonempty set. Suppose the mappingsatisfies (1)for allandif and only if; (2)for all; (3)for all. Then, is called a cone metric on, andis called a cone metric space (with Banach algebra).

For more details about cone metric spaces with Banach algebras, we refer the readers to [6].

*Definition 4 (see [8]). *Letbe a cone metric space, and let and be a sequence in. Then, (1)converges towhenever for each with there is a natural numbersuch that for all . We denote this by or ; (2)is a Cauchy sequence whenever for each with there is a natural number such that for all ; (3)is a complete cone metric space if every Cauchy sequence is convergent.

The following facts are often used.

Proposition 5 (see [8]). *Letbe a cone metric space, let be a normal cone with normal constant, and letbe a sequence in. Then, converges toif and only if.*

Proposition 6 (see [8]). *Letbe a cone metric space, let be a normal cone with normal constant, and letbe a sequence in. Then, is a Cauchy sequence if and only if .*

#### 2. Main Results

In this section we will define quasicontractions in the setting of cone metric spaces with Banach algebras and prove the fixed point theorem of such mappings.

*Definition 7. *Letbe a cone metric space with Banach algebra. A mappingis called a quasicontraction if for somewithand for all, one has
where

*Remark 8. *In Definition 7, we only suppose the spectral radius ofis less than 1, while neithernoris assumed. In fact, the conditionis weaker than that. See the example in [6].

Theorem 9. *Letbe a complete cone metric space with a Banach algebra, and let be a normal cone with normal constant. If the mappingis a quasicontraction, thenhas a unique fixed point in. And for any, iterative sequenceconverges to the fixed point.*

In the rest of the paper, we choose and denote. For the sake of clarity, we divide the proof into several steps.

Lemma 10. *Assume that the hypotheses in Theorem 9 are satisfied. Then, for each, and for allsuch that, one has
*

*Proof. *We present the proof by induction.

When, which implies, the conclusion is trivial.

Assume that the statement is true for; that is,
Now, we will prove that the statement is true for. Note that in this case, if, then the statement is just (8). Thus, without loss of generality, we suppose thatand and denote.

By the definition of quasicontraction, we have
where

Firstly, we consider the case that; that is,

If, then
and the statement follows.

If, then
and the statement also follows.

If, then we setand we have

If, then
which implies
Note that and thatandcommute. Multiplying both sides by, we have
and the statement also follows.

If, then
and the statement also follows.

Secondly, we consider the case that.

If or or, then, by (8), we have
and the statement follows.

If or, then we setor, respectively. And we have

In conclusion from discussions of both cases, it results that either the proof is complete, that is,
or there exists an integersuch that

As for the latter situation, we continue in a similar way, and come to the result that either
which implies that
and the proof is complete, or there exists an integersuch that
which implies that

Generally, if the procedure ends by the -th step with , that is, there exist integers
such that
and such that
then
Hence, the proof is complete.

Finally, if the procedure continues more thansteps, then there existintegers
such that
Thus, there must exist two integers,and, say, such that
From (32), one sees that
and therefore
Note that
which impliesis invertible. And since that
we have
So,

Therefore, by induction, the statement is proved.

*Remark 11. *Lemma 10 simply says that

Lemma 12. *Assume that the hypotheses in Theorem 9 are satisfied. Then, is a Cauchy sequence.*

*Proof. *For, denote that
By the definition of quasicontraction, it follows that, for each, there exists, such that
Consequently,
where
and the last inequality comes from Lemma 10.

By the normality of, and noting that, we have
The proof is complete.

Now, we finish the remaining part of the proof of Theorem 9.

*Proof. *By Lemma 12 and the completeness of, there issuch that. Then,
where

If or or, then. Hence,

If, then
Hence,

If, then
Hence,
as.

In each case, we have. Thus, .

Now, ifis another fixed point, then
where

If, then.

If, then
which implies

Thus, the fixed point is unique. And we obtain Theorem 9.

#### Acknowledgments

The authors are extremely grateful to the referees for their useful comments and suggestions. The research is partially supported by Doctoral Initial Foundation of Hanshan Normal University, China (no. QD20110920).

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#### Copyright

Copyright © 2013 Hao Liu and Shaoyuan Xu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.